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Let $X$ be a metric space, $A$, $B$ two compact subsets of $X$ such that the Hausdorff distance $dist_H$ between $A$ and $B$ is small: $$ dist_H(A, B) \leq \epsilon, $$ with $\epsilon > 0$. Does it imply that the Hausdorff dimensions $dim_H A$ and $dim_H B$ are closed to each other ? In other words, is the Hausdorff dimension continuous with respect to the Hausdorff distance ?

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Simple example. Let $n$ be a positive integer. Let $$ A = [0,1]\quad\text{and}\quad B_n = \left\{\frac{k}{n}\;:\;k=0,1,2,\dots,n\right\} $$ Then $\operatorname{dist}_\mathrm{H}(A,B_n) =\frac{1}{2n}$; $\dim_H A = 1$; $\dim_H B_n = 0$.

So $\operatorname{dist}_\mathrm{H}(A,B_n) \to 0$ but $\dim_H B_n \not\to \dim_H A$.

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  • $\begingroup$ Thank you for this nice example. $\endgroup$ – user776060 Apr 20 at 15:15
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No. For instance, if $(X,d)$ is locally compact and $K\subseteq X$ is compact, then for all $\varepsilon>0$ there is some relatively compact open set $U$ such that $K\subseteq U\subseteq K_\varepsilon$, and therefore $d_H\left(K,\overline U\right)\le \varepsilon$. However, $\dim_H\overline U$ is larger than the Hausdorff dimension of some open ball. And, say, in $\Bbb R^n$ the dimension all balls is $n$, while there are of course subsets of Hausdorff dimension smaller than $1$.

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  • $\begingroup$ Thank you very much for your answers. $\endgroup$ – user776060 Apr 20 at 14:28
  • $\begingroup$ What if $A$ and $B$ are connected sets? $\endgroup$ – user776060 Apr 20 at 14:28
  • $\begingroup$ @bill Irrelevant. In fact, if $K$ is connected and balls are connected, then $U$ may be chosen to be connected. $\endgroup$ – Gae. S. Apr 20 at 14:35
  • $\begingroup$ Thank you again. However, we have $\vert diam(A) - diam(B) \vert \leq dist_H(A, B)$. So, are there any other measures or dimensions of $A$ and $B$ whose difference is bounded by the Hausdorff distance? Maybe the topological dimension? $\endgroup$ – user776060 Apr 20 at 14:45
  • $\begingroup$ @bill Topological dimension won't do, because it's still $n$ for the $n$-ball. Others I don't know (which doesn't by any means exclude the possibility). $\endgroup$ – Gae. S. Apr 20 at 14:50

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